Past Exam Questions

(a) To form a committee with more men than women, consider the following cases:

1. 5 men and 0 women: \(\binom{8}{5} \times \binom{7}{0} = 56\)
2. 4 men and 1 woman: \(\binom{8}{4} \times \binom{7}{1} = 490\)
3. 3 men and 2 women: \(\binom{8}{3} \times \binom{7}{2} = 1176\)

Adding these gives the total number of ways: \(56 + 490 + 1176 = 1722\).

(b) To divide the 15 members into groups of 3, 5, and 7:

First, choose 3 people for the first group: \(\binom{15}{3}\).

Then, choose 5 people for the second group from the remaining 12: \(\binom{12}{5}\).

The remaining 7 people form the last group: \(\binom{7}{7} = 1\).

The total number of ways is \(\binom{15}{3} \times \binom{12}{5} = 455 \times 792 = 360360\).

We need to consider two cases: when Tom and Henry are both in the team and when both are not in the team.

Case 1: Tom and Henry are both in the team.

If Tom and Henry are in the team, we need to choose 1 more boy from the remaining 10 boys. This can be done in \(\binom{10}{1}\) ways. We also need to choose 3 girls from 9, which can be done in \(\binom{9}{3}\) ways.

The number of ways for this case is:

\(\binom{10}{1} \times \binom{9}{3} = 10 \times 84 = 840 \text{ ways}\)

Case 2: Tom and Henry are both not in the team.

If Tom and Henry are not in the team, we need to choose 3 boys from the remaining 10 boys. This can be done in \(\binom{10}{3}\) ways. We also need to choose 3 girls from 9, which can be done in \(\binom{9}{3}\) ways.

The number of ways for this case is:

\(\binom{10}{3} \times \binom{9}{3} = 120 \times 84 = 10080 \text{ ways}\)

Total number of ways:

Adding both cases together, we get:

\(840 + 10080 = 10920 \text{ ways}\)

To solve this problem, we need to distribute 9 pies among 3 people such that each person receives an odd number of pies. The possible distributions of odd numbers that sum to 9 are (1, 1, 7), (1, 3, 5), and (3, 3, 3).

1. For the distribution (1, 1, 7):

Choose 1 pie for the first person: \(\binom{9}{1}\)

Choose 1 pie for the second person from the remaining 8 pies: \(\binom{8}{1}\)

The third person gets the remaining 7 pies: \(\binom{7}{7}\)

Since the people are distinct, multiply by the number of ways to assign these distributions: \(\binom{3}{1}\)

Total ways for (1, 1, 7) = \(9 \times 8 \times 1 \times 3 = 216\)

2. For the distribution (1, 3, 5):

Choose 1 pie for the first person: \(\binom{9}{1}\)

Choose 3 pies for the second person from the remaining 8 pies: \(\binom{8}{3}\)

The third person gets the remaining 5 pies: \(\binom{5}{5}\)

Since the people are distinct, multiply by the number of ways to assign these distributions: \(3!\)

Total ways for (1, 3, 5) = \(9 \times 56 \times 1 \times 6 = 3024\)

3. For the distribution (3, 3, 3):

Choose 3 pies for the first person: \(\binom{9}{3}\)

Choose 3 pies for the second person from the remaining 6 pies: \(\binom{6}{3}\)

The third person gets the remaining 3 pies: \(\binom{3}{3}\)

Total ways for (3, 3, 3) = \(84 \times 20 \times 1 = 1680\)

Adding all the possibilities: \(216 + 3024 + 1680 = 4920\)

(i) There are 11 questions in total (8 from Part A and 3 from Part B). The number of ways to choose 6 questions from 11 is given by the combination formula:

\(\binom{11}{6} = 462\)

(ii) Candidates must choose at least 4 questions from Part A. We consider the following cases:

• 4 questions from Part A and 2 from Part B: \(\binom{8}{4} \times \binom{3}{2} = 70 \times 3 = 210\)
• 5 questions from Part A and 1 from Part B: \(\binom{8}{5} \times \binom{3}{1} = 56 \times 3 = 168\)
• 6 questions from Part A: \(\binom{8}{6} = 28\)

Adding these gives: \(210 + 168 + 28 = 406\)

(iii) Candidates must either choose both question 1 and question 2 in Part A, or choose neither. Consider the cases:

• Choose both question 1 and 2, then choose 4 more from the remaining 9 questions: \(\binom{9}{4} = 126\)
• Choose neither question 1 nor 2, then choose 6 from the remaining 9 questions: \(\binom{9}{6} = 84\)

Adding these gives: \(126 + 84 = 210\)

(i) If both boys are in the team, we need to choose 2 more people from the remaining 7 people (3 girls and 4 adults). The number of ways to choose 2 people from 7 is given by the combination formula: \(\binom{7}{2} = 21\).

(ii) If the adults are either all in the team or all not in the team, we consider two cases:

• All adults in the team: This is not possible since the team size is 4 and there are 4 adults.
• All adults not in the team: We choose all 4 members from the 5 non-adults (2 boys and 3 girls). The number of ways is \(\binom{5}{4} = 5\).

Thus, the total number of ways is \(1 + 5 = 6\).

(iii) If at least 2 girls are in the team, we consider two cases:

• 2 girls in the team: Choose 2 girls from 3 and 2 others from the remaining 6 people (2 boys and 4 adults). The number of ways is \(\binom{3}{2} \times \binom{6}{2} = 3 \times 15 = 45\).
• 3 girls in the team: Choose 3 girls from 3 and 1 other from the remaining 6 people. The number of ways is \(\binom{3}{3} \times \binom{6}{1} = 1 \times 6 = 6\).

Thus, the total number of ways is \(45 + 6 = 51\).

(ii) To find the number of ways to choose 6 images with 2 from each folder, we calculate:

\(\binom{3}{2} \times \binom{4}{2} \times \binom{8}{2} = 3 \times 6 \times 28 = 504\)

(iii) We need to consider different combinations of choosing images such that there are at least 3 from the friends folder and at least 1 from each of the other two folders. The possible distributions are:

1. 3 from Friends, 1 from Family, 2 from Holiday:

\(\binom{8}{3} \times \binom{3}{1} \times \binom{4}{2} = 56 \times 3 \times 6 = 1008\)

2. 3 from Friends, 2 from Family, 1 from Holiday:

\(\binom{8}{3} \times \binom{3}{2} \times \binom{4}{1} = 56 \times 3 \times 4 = 672\)

3. 4 from Friends, 1 from Family, 1 from Holiday:

\(\binom{8}{4} \times \binom{3}{1} \times \binom{4}{1} = 70 \times 3 \times 4 = 840\)

Summing these possibilities gives the total number of ways:

\(1008 + 672 + 840 = 2520\)

Let the number of geraniums be denoted by \(G\), roses by \(R\), and lilies by \(L\). We have the following conditions:

1. \(G + R + L = 25\)

2. \(G \geq 11\)

3. \(R = L\)

From condition 3, we have \(R = L\), so \(G + 2R = 25\).

We consider the possible values for \(G\):

1. \(G = 11\): Then \(11 + 2R = 25\) gives \(R = 7\) and \(L = 7\). The number of ways to choose the flowers is \(\binom{15}{11} \times \binom{10}{7} \times \binom{8}{7} = 1310400\).
2. \(G = 13\): Then \(13 + 2R = 25\) gives \(R = 6\) and \(L = 6\). The number of ways to choose the flowers is \(\binom{15}{13} \times \binom{10}{6} \times \binom{8}{6} = 617400\).
3. \(G = 15\): Then \(15 + 2R = 25\) gives \(R = 5\) and \(L = 5\). The number of ways to choose the flowers is \(\binom{15}{15} \times \binom{10}{5} \times \binom{8}{5} = 14112\).

Adding these possibilities gives the total number of selections:

\(1310400 + 617400 + 14112 = 1941912\).

The word HAPPINESS consists of the letters H, A, P, P, I, N, E, S, S. We need to select 4 letters with no Ps and either one or two Ss.

First, remove the Ps, leaving us with the letters H, A, I, N, E, S, S.

Case 1: One S

Select 1 S and 3 other letters from H, A, I, N, E. This can be done in \(\binom{5}{3} = 10\) ways.

Case 2: Two Ss

Select 2 Ss and 2 other letters from H, A, I, N, E. This can be done in \(\binom{5}{2} = 10\) ways.

\(Total number of selections = 10 + 10 = 20.\)

Alternatively, select 3 letters from H, A, I, N, E, S (6 letters total) and ensure at least one S is included. This can be done in \(\binom{6}{3} = 20\) ways.

(i) Consider the different combinations of players:

• 5 batsmen, 5 bowlers, 1 wicketkeeper: \(\binom{10}{5} \times \binom{9}{5} \times \binom{2}{1} = 63504\)
• 5 batsmen, 4 bowlers, 2 wicketkeepers: \(\binom{10}{5} \times \binom{9}{4} \times \binom{2}{2} = 31752\)
• 6 batsmen, 4 bowlers, 1 wicketkeeper: \(\binom{10}{6} \times \binom{9}{4} \times \binom{2}{1} = 52920\)

Total ways: \(63504 + 31752 + 52920 = 148176\)

(ii) The number of different arrangements of the presents is given by:

\(\frac{11!}{5!4!2!} = 6930\)

(iii) Consider omitting one type of present:

• Omit a pen: \(\frac{10!}{4!4!2!} = 3150\)
• Omit a diary: \(\frac{10!}{5!3!2!} = 2520\)
• Omit a notebook: \(\frac{10!}{5!4!} = 1260\)

Total arrangements: \(3150 + 2520 + 1260 = 6930\)

(i) The committee must have either 4 men and 2 women or 5 men and 1 woman. The number of ways to choose 4 men from 10 and 2 women from 9 is \(\binom{10}{4} \times \binom{9}{2}\). The number of ways to choose 5 men from 10 and 1 woman from 9 is \(\binom{10}{5} \times \binom{9}{1}\). Therefore, the total number of committees is:

\(\binom{10}{4} \times \binom{9}{2} + \binom{10}{5} \times \binom{9}{1} = 210 \times 36 + 252 \times 9 = 7560 + 2268 = 9828\).

(ii) To find the probability that Albert and Tracey are both on the committee, we first choose 3 more men from the remaining 9 men and 1 more woman from the remaining 8 women. The number of ways to do this is \(\binom{9}{3} \times \binom{8}{1}\). The total number of committees is 9828, so the probability is:

\(\frac{\binom{9}{3} \times \binom{8}{1}}{9828} = \frac{84 \times 8}{9828} = \frac{672}{9828} = 0.0812\).

(iii) To find the number of committees that include either Albert or Tracey but not both, we calculate separately:

- Albert but not Tracey: Choose 3 more men from 9 and 2 women from 8: \(\binom{9}{3} \times \binom{8}{2}\).

- Tracey but not Albert: Choose 4 men from 9 and 1 more woman from 8: \(\binom{9}{4} \times \binom{8}{1}\).

The total number of such committees is:

\(\binom{9}{3} \times \binom{8}{2} + \binom{9}{4} \times \binom{8}{1} = 3360 + 1134 = 4494\).

(iv) The committee consists of 4 men and 2 women. The total number of ways to arrange 6 people is \(6!\). The number of ways to arrange the 4 men and 2 women such that the women are not next to each other is:

- Total arrangements: \(6! = 720\).

- Arrangements with women together: Treat the women as a single unit, so \(5! \times 2! = 240\).

- Arrangements with women not together: \(720 - 240 = 480\).

The probability is:

\(\frac{480}{720} = \frac{2}{3}\).

(iii) To use 4 different colours, choose 4 colours from the 6 available. This can be done in \(\binom{6}{4}\) ways. For each selection, the 4 pegs can be arranged in \(4!\) ways. Thus, the total number of arrangements is \(4! \times \binom{6}{4} = 360\).

(iv) To use 3 different colours, consider cases where two pegs are of one colour and the other two are of different colours. For example, 2 red, 1 blue, 1 green. The number of arrangements for each case is \(\frac{4!}{2!}\). There are \(\binom{6}{3}\) ways to choose 3 colours from 6. Thus, the total number of arrangements is \(3 \times \frac{4!}{2!} \times \binom{6}{3} = 720\).

(v) Using any of her 12 pegs, consider cases where two pegs are of one colour and two pegs are of another colour. For example, 2 red, 2 blue. The number of arrangements for each case is \(\frac{4!}{2!2!} = 6\). There are \(\binom{6}{2}\) ways to choose 2 colours from 6. Thus, the total number of arrangements is \(6 \times \binom{6}{2} = 90\).

Adding all these possibilities gives the total number of arrangements: \(360 + 720 + 90 = 1170\).

(a) To divide the remaining 12 members into cars of 6, 5, and 4 people, we calculate:

\(\binom{12}{6} \times \binom{6}{5} \times \binom{1}{1} = 924 \times 6 \times 1 = 5544\)

However, considering the arrangement of cars, the correct calculation is:

\(\frac{12!}{6! \times 5! \times 4!} = 27720\)

(c) To choose a team of 4 adults and 3 children from different families:

Choose 4 adults from 7: \(\binom{7}{4} = 35\)

Choose 3 children from different families: \(3! = 6\)

Total ways: \(35 \times 6 = 210\)

However, the correct calculation is:

\(\binom{7}{4} \times \binom{3}{1} \times \binom{3}{1} \times \binom{3}{1} = 420\)

(d) To ensure at least one of Mr Kenny or Mr Lizo is included:

Total ways without restriction: \(\binom{7}{4} \times \binom{8}{3} = 1960\)

Ways excluding both Mr Kenny and Mr Lizo: \(\binom{5}{4} \times \binom{8}{3} = 280\)

Thus, ways including at least one: \(1960 - 280 = 1680\)

To solve this problem, we need to determine the number of ways to distribute 10 different mugs between Lucy and Monica such that both receive an odd number of mugs.

Since the total number of mugs is 10, the possible distributions where both receive an odd number are:

• Lucy receives 1 mug and Monica receives 9 mugs.
• Lucy receives 3 mugs and Monica receives 7 mugs.
• Lucy receives 5 mugs and Monica receives 5 mugs.
• Lucy receives 7 mugs and Monica receives 3 mugs.
• Lucy receives 9 mugs and Monica receives 1 mug.

For each distribution, we calculate the number of combinations:

\(\binom{10}{1} + \binom{10}{3} + \binom{10}{5} + \binom{10}{7} + \binom{10}{9}\)

Calculating each term:

• \(\binom{10}{1} = 10\)
• \(\binom{10}{3} = 120\)
• \(\binom{10}{5} = 252\)
• \(\binom{10}{7} = 120\)
• \(\binom{10}{9} = 10\)

Summing these values gives:

\(10 + 120 + 252 + 120 + 10 = 512\)

Thus, there are 512 different ways to distribute the mugs such that both Lucy and Monica receive an odd number of mugs.

(i) To have more girls than boys in the team of 5, the possible combinations are:

• 5 girls and 0 boys: \(\binom{8}{5} \times \binom{6}{0} = 56\)
• 4 girls and 1 boy: \(\binom{8}{4} \times \binom{6}{1} = 420\)
• 3 girls and 2 boys: \(\binom{8}{3} \times \binom{6}{2} = 840\)

\(Total = 56 + 420 + 840 = 1316\)

(ii) If three of the boys are cousins and are either all in the team or all not in the team, consider two cases:

• All cousins in the team: Choose 2 more from the remaining 11 (3 boys + 8 girls): \(\binom{11}{2} = 55\)
• All cousins not in the team: Choose 5 from the remaining 11 (3 boys + 8 girls): \(\binom{11}{5} = 462\)

\(Total = 55 + 462 = 517\)

To find the number of ways to choose the group, we use combinations for each voice part:

1. Choose 10 sopranos from 13: \(\binom{13}{10}\)

2. Choose 9 altos from 12: \(\binom{12}{9}\)

3. Choose 4 tenors from 6: \(\binom{6}{4}\)

4. Choose 4 basses from 7: \(\binom{7}{4}\)

The total number of ways to choose the group is the product of these combinations:

\(\binom{13}{10} \times \binom{12}{9} \times \binom{6}{4} \times \binom{7}{4}\)

Calculating each combination:

\(\binom{13}{10} = 286\)

\(\binom{12}{9} = 220\)

\(\binom{6}{4} = 15\)

\(\binom{7}{4} = 35\)

Thus, the total number of ways is:

\(286 \times 220 \times 15 \times 35 = 33,033,000\)

To select the houses, we need to choose:

1. 1 house from the 2 houses in style B: \(\binom{2}{1} = 2\) ways.
2. 1 house from the 3 houses in style C: \(\binom{3}{1} = 3\) ways.
3. 2 more houses from the remaining 7 houses (2 in style A, 4 in style D, 1 in style E): \(\binom{7}{2} = 21\) ways.

Thus, the total number of ways to select the houses is:

\(2 \times 3 \times 21 = 126\)

To find the number of different possible selections, we use combinations to select the CDs from each category.

First, select 2 pop music CDs from 6: \\(_6C_2 = \frac{6!}{2!(6-2)!} = 15\\)

Next, select 2 jazz CDs from 3: \\(_3C_2 = \frac{3!}{2!(3-2)!} = 3\\)

Finally, select 1 classical CD from 2: \\(_2C_1 = \frac{2!}{1!(2-1)!} = 2\\)

Multiply the combinations together to find the total number of selections: \\(15 \times 3 \times 2 = 90\\)

The word REFRIGERATOR consists of the letters: R, E, F, R, I, G, E, R, A, T, O, R.

We need to select four letters with no Rs and exactly two Es.

First, remove all Rs from the word, leaving: E, F, I, G, E, A, T, O.

Now, we have 8 letters: E, E, F, I, G, A, T, O.

We need to select 4 letters, including exactly 2 Es.

\(Choose 2 Es: inom{2}{2} = 1.\)

\(Choose 2 more letters from the remaining 6 letters (F, I, G, A, T, O): inom{6}{2} = 15.\)

\(Thus, the total number of selections is: 1 imes 15 = 15.\)

To solve this problem, we need to consider the different cases where at least one woman is included in the selection of three people.

1. Case 1: 1 woman and 2 men are chosen.

   The number of ways to choose 1 woman from 3 is given by the combination \(\binom{3}{1} = 3\).

   The number of ways to choose 2 men from 6 is given by the combination \(\binom{6}{2} = 15\).

   Thus, the total number of ways for this case is \(3 \times 15 = 45\).

2. Case 2: 2 women and 1 man are chosen.

   The number of ways to choose 2 women from 3 is given by the combination \(\binom{3}{2} = 3\).

   The number of ways to choose 1 man from 6 is given by the combination \(\binom{6}{1} = 6\).

   Thus, the total number of ways for this case is \(3 \times 6 = 18\).

3. Case 3: 3 women are chosen.

   The number of ways to choose 3 women from 3 is given by the combination \(\binom{3}{3} = 1\).

Adding all these cases together, the total number of ways to choose at least one woman is \(45 + 18 + 1 = 64\).

Alternatively, calculate the total number of ways to choose any 3 people from 9 (6 men + 3 women) without restriction: \(\binom{9}{3} = 84\).

Subtract the number of ways to choose only men (no women): \(\binom{6}{3} = 20\).

Thus, the number of ways to choose at least one woman is \(84 - 20 = 64\).

To solve this problem, we need to select 2 married couples and 1 additional person from the remaining passengers.

First, choose 2 married couples from the 3 available couples. The number of ways to do this is given by the combination formula:

\(\binom{3}{2} = 3\)

After choosing 2 married couples, there are 7 passengers left (5 unmarried and 2 from the remaining couple). We need to choose 1 additional person from these 7 passengers:

\(\binom{7}{1} = 7\)

Therefore, the total number of ways to choose the passengers is:

\(3 \times 7 = 21\)